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The Mordell-Weil theorem

MA 254 notes: Diophantine Geometry(Distilled from [Hindry-Silverman], [Manin], [Serre])

Dan Abramovich

Brown University

Janoary 27, 2016

Abramovich MA 254 notes: Diophantine Geometry 1 / 16


Mordell Weil and Weak Mordell-Weil

Theorem (The Mordell-Weil theorem)

Let A/k be an abelian variety over a number field. Then A(k) isfinitely generated.

We deduce this from

Theorem (The weak Mordell-Weil theorem)

Let A/k be as above and m ≥ 2. Then A(k)/mA(k) is finite.



Mordell-Weil follows from Weak Mordell-Weil

The reduction comes from the following.

Lemma (Infinite Descent Lemma)

Let G be an abelian group, q : G → R a quadratic form. Assume(i) for all C ∈ R the set {x ∈ G |q(x) ≤ C} is finite, and(ii) there is m ≥ 2 such that G/mG is finite.

Then G is finitely generated. In fact if {gi} are representatives forG/mG and C0 = max{q(gi )} then the finite setS := {x ∈ G |q(x) ≤ C0} generates G .

indeed the canonical Néron-Tate height ĥD on A(k) associated toan ample symmetric D gives a quadratic form satisfying (i), and(ii) follows from Weak Mordell-Weil. The Mordell-Weil theoremfollows. ♠



Proof of the Infinite Descent Lemma for m ≥ 3

q(x) ≥ 0 otherwise the elements {nx} would violate (i).Write |x | =

√q(x), c0 = max{|gi |},

S√m

kc0

= {x ∈ G : |x | ≤√

mk

c0}.

Clearly G = ∪S√m

kc0

, so enough to show S√m

kc0⊂ 〈S〉 for all

k.

We use induction on k ≥ 0. Clearly Sc0 = S ⊂ 〈S〉.Assume S√

mkc0⊂ 〈S〉 and let x ∈ S√

mk+1

c0r S√

mkc0

.

x ∈ gi + mG for some i so there is x1 such thatm x1 = x − gi . So

|x1| =|x − gi |

m≤ |x |+ |gi |

m≤√

mk+1

+ 1

mc0 ≤

√m

kc0

since m ≥ 3. ♠Abramovich MA 254 notes: Diophantine Geometry 4 / 16


Proof of Weak Mordell-Weil - Finite Kernel Lemma

Lemma (Finite Kernel Lemma)

Fix a finite k ′/k. Then Ker(A(k)/mA(k) → A(k ′)/mA(k ′)

)is finite.

Proof of lemma

Note this kernel is Bk ′/k := (A(k) ∩mA(k ′))/mA(k).Since for a further extension k ′′/k ′ we have Bk ′/k ⊂ Bk ′′/k , wemay replace k ′ by a further extension.

In particular we may assume k ′/k Galois, with Galois groupG . The lemma now follows from the following, since G andAm(k̄) are finite:

Sublemma

There is an injective function Bk ′/k → Maps(G ,Am(k̄)).



Proof of Weak Mordell-Weil - reduction - sublemma

Proof of sublemma

For x ∈ Bk ′/k fix y ∈ A(k ′) such that [m]y represents x .Consider the function fx : G → A(k ′) where f (σ) = yσ − y .Note [m]fx(σ) = [m](y

σ − y) = [my ]σ − [m]y = xσ − x = 0.This defines Bk ′/k → Maps(G ,Am(k̄)).We claim it is injective.

Assume fx = fx ′ , and let y , y′ be the chosen points in A(k ′).

So y ′σ − y ′ = yσ − y , hence (y ′ − y)σ = y ′ − y for all σ.So y ′ − y ∈ A(k) and [m](y ′ − y) ∈ mA(k).Hence x − x ′ = 0 in Bk ′/k , as needed. ♠♠



Theorem (Chevalley-Weil+Hermite for [m])

Fix an abelian variety A/k and an integer m. There is a finiteextension k ′/k such that if x ∈ A(k) then there is y ∈ A(k ′) suchthat [m]y = x.

Proof of Weak Mordell-Weil assuming Chevalley-Weil+Hermite

Let k ′ be as in Chevalley-Weil+Hermite for [m]. Then the mapA(k)/mA(k)→ A(k ′)/mA(k ′) is zero, hence by the Finite KernelLemma A(k)/mA(k) is finite. ♠

We will present two proofs of Chevalley-Weil+Hermite: one quickand dirty using Scheme Theory. One going through with rings anddifferentials explicitly.



Hermite’s theorem(s)

The following classical theorem appears in a first course:

Theorem (Hermite 1)

For any real B there are only finitely many number fields k with|dk | ≤ B. ♠

This one might not be as visible:

Theorem (Hermite 2)

For any integer n and finite set of primes S there are only finitelymany number fields k unramified outside S with [k : Q] ≤ n.

Hermite 2 follows from Hermite 1 given the following:

Proposition (Discriminant bound (Serre), not proven here)

Write [k : Q] = n and N =∏

p|dk p. Then |dk | ≤ (N · n)n.



Why number-theorists want to understand schemes

What I stated as “Chevalley-Weil+Hermite” is a consequence of

Proposition (Spreading out)

Let φ : Y → X be a finite unramified morphism of projectivevarieties, all over k. Then there is a finite set of primes S,projective schemes X → SpecOk,S , Y → SpecOk,S and a finiteunramified morphism ϕ : Y → X , whose restriction to Spec k isφ : Y → X .



Spreading out: Picture with point



Spreading out implies Chevalley-Weil

The reduction works as follows:

In our case take X = Y = A and φ = [m].

Since X → SpecOk,S pojective, a point x ∈ X (k) extends toa morphism x̃ : SpecOk,S → X .Write Yx = SpecOk,S ×X Y.Since Y → X is unramified we have ΩY/X = 0.ΩYx/SpecOk,S is the pullback of this to Yx , so also 0.If [m]y = x then y ⊂ Yx and its closure ỹ is unramified over x̃ .This means that O(ỹ) is finite unramified over Ok,S , sok(y)/k unramified away from S .

By Hermite 2 there are only finitely many such k(S). Let k ′

be the Galois closure of their compositum.

so y ∈ A(k ′). ♠



Exorcising schemes: Spreading Out explained

Step 1: coordinates

Say Y ∈ Pm and X ∈ Pn. We may replace Y ⊂ Pm by thegraph of φ.

We now have Y ⊂ Pm × Pn, and φ is the restriction of thenatural projection Pm × Pn → Pn.For each coordinate Xi of Pn we have Ui = X r Z (Xi ) affine,with coordinate ring Ai = k[x0, . . . ,��ZZxi , . . . , xn]/(f1i , . . . , fri i ).

“φ finite” means: preimage of affine is affine, correspondingto a finite ring extension.

So Vi = φ−1Ui affine, with ring Bi finite over Ai . Let

{y1i , . . . , yki i} be module generators and gji the modulerelations and YjiYj ′i = hjj ′i the ring relations, with gji , hjj ′iAi -linear in Y1i , . . . ,Yki i :

Bi = Ai [Y1i , . . . ,Yki i ]/({gji}, {YjiYj ′i − hjj ′i}).




Step 2: shrinking to define rings and maintainfiniteness

There are finitely many nonzero coefficients {aα} infj ,i , gji , hjj ′i , giving a subring Ok,S0 := Ok [{aα, a−1α }] of k inwhich aα are units. Here S0 is the set of places appearing inthe factorizations of the aα.

Let Ai = Ok,S [x0, . . . ,��ZZxi , . . . , xn]/(f1i , . . . , fri i ) andBi = Ai [Y1i , . . . ,Yki i ]/({gji}, {YjiYj ′i − hjj ′i}).We clearly have Ai = Ai ⊗Ok,S k and Bi = Bi ⊗Ok,S k .By construction Ai 〈Yji 〉 → Bi is a surjective modulehomomorphism.

So Bi is a finite Ai -algebra.




Step 3: shrinking to evade ramification

The statement “Y → X unramified” is equivalent to“Vi → Ui unramified”.This is equivalent to “ΩBi/Ai = 0”.

Consider the finitely generated Bi -module ΩBi/Ai .We have ΩBi/Ai ⊗Ok,S k = ΩBi/Ai = 0.So AnnOk,S (ΩBi/Ai ) 6= 0. In other words there is nonzeroc ∈ Ok,S such that cΩBi/Ai = 0.Replacing Ok,S by Ok,S [c−1] ⊂ k we may assume ΩBi/Ai = 0,hence Bi is an unramified Ai -algebra.

In the language of spreading out, Y := ∪SpecBi andX := ∪SpecAi . ♠(Spreading Out)



Exorcising schemes: Chevalley-Weil explained

Proposition

Let x ∈ X (k) and φ(y) = x. Then k(y)/k is unramified outside S.

Fix a nonzero prime p ⊂ Ok,S . We show k(y)/k is unramifiedat p.

Let x = (α0, . . . , αn) ∈ X . Since Ok,p is principal we can takeαi ∈ Ok,p relatively prime.Without loss of generality the uniformizer πp 6 |α0.Replacing αi by αi/α0 we may assume α0 = 1.

Consider the epimorphism A0 → k given by the maximal ideal(x1 − α1, . . . , xn − αn).

It gives (A0)px̃−→ Ok,p (the image is no bigger).



Exorcising schemes: Chevalley-Weil explained

Proof of the proposition

Consider C := B0 ⊗A0 Ok,p.Since A0 → B0 is finite and unramified, also Ok,S → C isfinite and unramified.

Consider the commutative diagram

B0 //

**

C

**A0

OO

//

**

Ok,p

OO

**

B0 // k(y)

A0

OO

// k

OO

The universal property of tensor gives an arrow C → k(y).Its image is a subring Ry ,p ⊂ k(y) finite unramified over Ok,p,which must be Ok(y),p. ♠



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